Find a Lyapunov function of the form $ax^2+by^2+cz^2$

Find a Lyapunov function of the form $ax^2+by^2+cz^2$ to determine whether the equilibrium points of the differential equation:

$\dot{x}=2y(z-1)$

$\dot{y}=-x(z-1)$

$\dot{z}=-z^3$

are stable, asymptotically stable or unstable.

I don't really know if what I did is correct, and if it isn't where is the error:

By Lyapunov's theorem, the function $V(x,y,z)=ax^2+by^2+cz^2$ has to have a local minimum in the equilibrium point of the differential equation (which is $(0,0,0)$), and $\dot{V}(x,y,z)$ has to be either $\leq0$ (stable) or $<0$ (asymptotically stable) in a neighbourhood of $(0,0,0)$.

So, $\dot{V}(x,y,z)=2ax\dot{x}+2by\dot{y}+2cz\dot{z}=2ax[2y(z-1)]+2by[-x(z-1)]-2cz^4 \leq0 \iff xyz(4a-2b)-xy(4a-2b)-2cz^4 \leq 0 \iff b=2a, c>0$

So, if I choose let's say, $a=1 \Rightarrow b=2$ and $c=1$, I end up with the Lyapunov's function $V(x,y,z)=x^2+2y^2+z^2$, which has a local minimum in $(0,0,0)$ and $\dot{V}<0$ (in a neighborhood not containing $(0,0,0)$), therefore the equilibrium point is asymptotically stable.

My doubt is whether is this correct, as I never took into account that $(0,0,0)$ has to be minimum, it just happened to verify the function.

Solution 1:

Yes, this is partially correct. Note that you found a condition to $\dot{V}\le0$ so, as you said, origin is stable (not asymptotically stable). Note also that this system has a nice geometric description: All (non-constant) solutions are confined in cylinders $2y^2+x^2=C$. Also all solutions with $z(0)\neq0$ has $z\to0$. So we have helix curves foliating all $R^3$ above and below the $z=0$ plane.